How do you solve Range Addition II on LeetCode?

Range Addition II (LeetCode #598) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(k) time complexity and O(1) space complexity. Key insight: The maximum integer in the matrix is determined by the smallest dimensions affected by all operations.

What is the time complexity of Range Addition II?

The optimal solution for Range Addition II runs in O(k) time and uses O(1) space. The time complexity is O(k) where k is the number of operations, which is efficient compared to the brute-force approach.

What is the brute force solution for Range Addition II?

The brute force approach for Range Addition II is Brute Force, with O(n²) time complexity and O(1) space complexity. We can create the matrix and apply each operation directly. This approach is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(k).

What algorithm or data structure is used to solve Range Addition II?

Range Addition II uses the following concepts: Array, Math. The recommended patterns to study are: Array, Math.

What are the interview tips for Range Addition II?

Always clarify the problem and constraints before jumping into coding. Think about the efficiency of your solution and whether a brute-force approach is necessary. Practice dry-running your code with sample inputs to catch errors early.

What are common mistakes when solving Range Addition II?

Students often try to implement the brute-force solution without realizing its inefficiency. Failing to consider edge cases, such as when ops is empty.

#598

Range Addition II

Easy

ArrayMathArrayMath

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(k)
Space	O(1)	O(1)

💡

Intuition

Time O(k)Space O(1)

Instead of modifying the matrix, we can track the minimum dimensions affected by all operations. The maximum integer will be determined by the smallest dimensions, and we can calculate the count directly.

⚙️

Algorithm

3 steps

1Step 1: Initialize min_row and min_col to m and n respectively.
2Step 2: For each operation in ops, update min_row and min_col to the minimum of their current values and the operation's dimensions.
3Step 3: The maximum integer in the matrix is the number of operations applied, which is equal to the product of min_row and min_col.

solution.py8 lines

1def maxCount(m, n, ops):
2    if not ops:
3        return m * n
4    min_row, min_col = m, n
5    for a, b in ops:
6        min_row = min(min_row, a)
7        min_col = min(min_col, b)
8    return min_row * min_col

ℹ

Complexity note: The time complexity is O(k) where k is the number of operations, which is efficient compared to the brute-force approach.

1The maximum integer in the matrix is determined by the smallest dimensions affected by all operations.
2If no operations are performed, the entire matrix is filled with zeros.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.